The finalists in an essay competition are Lisa (L), Melina (M), Ben (B), Danny (D), Eric (E), and Joan (J). Consider these finalists to be a population of interest. The possible samples (without Replacement)of size two that can be obtained from this population of six finalists are as follows. L,M L,B L,D L,E L,J M,B M,D
M,E M,J B,D B,E B,J D,E D,J E,J
If a simple random sampling method is used to obtain a sample of two of the finalists, what are the chances of selecting Lisa and Danny?
A)$\frac { 1 } { 6 }$
B) $\frac { 2 } { 15 }$
C) $\frac { 1 } { 15 }$
D) $\frac { 1 } { 3 }$

Question

The finalists in an essay competition are Lisa (L), Melina (M), Ben (B), Danny (D), Eric (E), and Joan (J). Consider these finalists to be a population of interest. The possible samples (without Replacement)of size two that can be obtained from this population of six finalists are as follows. L,M L,B L,D L,E L,J M,B M,D
M,E M,J B,D B,E B,J D,E D,J E,J
If a simple random sampling method is used to obtain a sample of two of the finalists, what are the chances of selecting Lisa and Danny?  
A)$\frac { 1 } { 6 }$
B) $\frac { 2 } { 15 }$
C) $\frac { 1 } { 15 }$
D) $\frac { 1 } { 3 }$

Quizplus · Accepted Answer

The total number of possible samples of size two from a population of six is calculated using the combination formula $C(n, k) = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items (in this case, 6 finalists) and $k$ is the size of each sample (in this case, 2). This gives us $C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2} = 15$. Since there is only one way to select Lisa and Danny as a pair, the probability is $\frac{1}{15}$.

The Finalists in an Essay Competition Are Lisa (L), Melina 16\frac { 1 } { 6 }61​

The Finalists in an Essay Competition Are Lisa (L), Melina $\frac { 1 } { 6 }$